integration seems gaussian, but can't solve

Question

i need some help in this integral

$$\int_{-\infty}^{\infty}\exp(-bx^2) \frac{d^2}{dx^2} \left(\exp(-bx^2)\right) dx$$

I tried differentiating $\displaystyle e^{-bx^2}$ twice and it came up weird , is there any other way to do it ?

Answer 1

Hint: Integration by parts. Let $f(x) = e^{-bx^2}$. Then the given integral is $$I = \int_{x=-\infty}^\infty f(x) f''(x) \, dx.$$ This suggests letting $u = f(x)$, $du = f'(x) \, dx$, $dv = f''(x) \, dx$, $v = f'(x)$, to give $$I = \left[ f(x) f'(x) \right]_{x=-\infty}^\infty - \int_{x=-\infty}^\infty (f'(x))^2 \, dx.$$ Now compute the appropriate limits and show that the first term is $0$, and then expand out $(f'(x))^2$. Then compute this integral using another integration by parts, with the choice $u = x$. This will result in one final integral of the form $$\int C e^{-Bx^2} \, dx$$ for some suitable constants $B, C$, which is a Gaussian integral.